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# 1-)-Write a cfg (contextfree grammar) for L - {a^n b^m c^k - k n + m.docx

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# 1-)-Write a cfg (contextfree grammar) for L - {a^n b^m c^k - k n + m.docx

1.).Write a cfg (contextÂfree grammar) for L = {a^n b^m c^k : k n + m n, m, k 0}.
2.).Write a cfg for L = {a^n b^n c^k : n 0, k 3}.
3).Write the grammar in #2 in CNF format.
Solution
Let G be the grammar with productions:
S EcC | aAE | AU A aA | B bB | C cC |
E aEc | F
F bFc |
U aUc | V
V bVc | bB
Note: L(E) = {an bmc k : k = n + m } (from 4(a) in Linz book) Let L1 = {an bmc k : k < n + m }, L2 = {an bmc k : k > n + m }
Claim: L(G) = L = L1 L2
Proof: Consider the leftmost productions following S Ã† EcC S EcC * an bmc (n+m)cC * an bmc (n+m)cck C an bmc (n+m)cck = an bmc (n+m+k+1) (where the first * follows from 4(a), the second from k repetitions of { C cC }) Since (n+m) < (n+m+k+1), L(EcC) L2 L Consider the leftmost productions following S aAE S aAE * aak AE aa k E * aak a n bmc (n+m)c = a(n+k+1)bmc (n+m) (where the first * follows from k repetitions of { A aA }, the second from 4(a)) Since (n+m+k+1) > (n+m), L(aAE) L1 L Consider the leftmost productions following S AU S AU * ak AU a k U * ak a n Ucn a k a n Vcn * ak a n bmVcmc n a k a n bmbBcmc n * ak a n bmbbj Bcmc n ak a n bmbbj c mc n = a(k+n)b(m+j+1)c (m+n) (where the first, second, third and fourth * are k, n, m, and j repetitions of { A aA }, { U aUc }, { V bVc }, and { B bB }, respectively)
Since ((k+n)+(m+j+1)) > (m+n), L(AU) L1 L L(G) = L(S) = L(EcC) L(aAE) L(AU) L1 L2 = L Consider w L1, w = an bmc (n+m+k) for some n, m, k, with k > 0 Then w has the leftmost derivation: S EcC * an bmc (n+m)cC * an bmc (n+m)cc(k-1)C an bmc (n+m)cck(k-1) = a n bmc (n+m+k) = w L2 L(G) Consider w L2, w = a(n+j)b(m+k)c (n+m) for some j, k, m, n, with j+k > 0 If k = 0, w has the leftmost derivation: S aAE * aa(j-1)AE aa(j-1)E * aa(j-1)a n Ecn aa(j-1)a n Fcn * aa(j- 1)a n bmFcmc n aa(j-1)a n bmc mc n = a(j+n)b(0+m)c (m+n) = w L1 (k = 0) L(G) If k > 0, then w has the leftmost derivation: S AU * aj AU aj U * aj a n Ucn aj a n Vcn * aj a n bmVcmc n a j a n bmbBcmc n aj a n bmbb(k-1)Bcmc n aj a n bmbb(k-1)c mc n = a(j+n)b(m+k)c (m+n) = w
L1 (k > 0) L(G) L1 L(G)
L = L1 L2 L(G)
L = L(G)
.

1.).Write a cfg (contextÂfree grammar) for L = {a^n b^m c^k : k n + m n, m, k 0}.
2.).Write a cfg for L = {a^n b^n c^k : n 0, k 3}.
3).Write the grammar in #2 in CNF format.
Solution
Let G be the grammar with productions:
S EcC | aAE | AU A aA | B bB | C cC |
E aEc | F
F bFc |
U aUc | V
V bVc | bB
Note: L(E) = {an bmc k : k = n + m } (from 4(a) in Linz book) Let L1 = {an bmc k : k < n + m }, L2 = {an bmc k : k > n + m }
Claim: L(G) = L = L1 L2
Proof: Consider the leftmost productions following S Ã† EcC S EcC * an bmc (n+m)cC * an bmc (n+m)cck C an bmc (n+m)cck = an bmc (n+m+k+1) (where the first * follows from 4(a), the second from k repetitions of { C cC }) Since (n+m) < (n+m+k+1), L(EcC) L2 L Consider the leftmost productions following S aAE S aAE * aak AE aa k E * aak a n bmc (n+m)c = a(n+k+1)bmc (n+m) (where the first * follows from k repetitions of { A aA }, the second from 4(a)) Since (n+m+k+1) > (n+m), L(aAE) L1 L Consider the leftmost productions following S AU S AU * ak AU a k U * ak a n Ucn a k a n Vcn * ak a n bmVcmc n a k a n bmbBcmc n * ak a n bmbbj Bcmc n ak a n bmbbj c mc n = a(k+n)b(m+j+1)c (m+n) (where the first, second, third and fourth * are k, n, m, and j repetitions of { A aA }, { U aUc }, { V bVc }, and { B bB }, respectively)
Since ((k+n)+(m+j+1)) > (m+n), L(AU) L1 L L(G) = L(S) = L(EcC) L(aAE) L(AU) L1 L2 = L Consider w L1, w = an bmc (n+m+k) for some n, m, k, with k > 0 Then w has the leftmost derivation: S EcC * an bmc (n+m)cC * an bmc (n+m)cc(k-1)C an bmc (n+m)cck(k-1) = a n bmc (n+m+k) = w L2 L(G) Consider w L2, w = a(n+j)b(m+k)c (n+m) for some j, k, m, n, with j+k > 0 If k = 0, w has the leftmost derivation: S aAE * aa(j-1)AE aa(j-1)E * aa(j-1)a n Ecn aa(j-1)a n Fcn * aa(j- 1)a n bmFcmc n aa(j-1)a n bmc mc n = a(j+n)b(0+m)c (m+n) = w L1 (k = 0) L(G) If k > 0, then w has the leftmost derivation: S AU * aj AU aj U * aj a n Ucn aj a n Vcn * aj a n bmVcmc n a j a n bmbBcmc n aj a n bmbb(k-1)Bcmc n aj a n bmbb(k-1)c mc n = a(j+n)b(m+k)c (m+n) = w
L1 (k > 0) L(G) L1 L(G)
L = L1 L2 L(G)
L = L(G)
.

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### 1-)-Write a cfg (contextfree grammar) for L - {a^n b^m c^k - k n + m.docx

1. 1. 1.).Write a cfg (contextÂfree grammar) for L = {a^n b^m c^k : k n + m n, m, k 0}. 2.).Write a cfg for L = {a^n b^n c^k : n 0, k 3}. 3).Write the grammar in #2 in CNF format. Solution Let G be the grammar with productions: S EcC | aAE | AU A aA | B bB | C cC | E aEc | F F bFc | U aUc | V V bVc | bB Note: L(E) = {an bmc k : k = n + m } (from 4(a) in Linz book) Let L1 = {an bmc k : k < n + m }, L2 = {an bmc k : k > n + m } Claim: L(G) = L = L1 L2 Proof: Consider the leftmost productions following S Ã† EcC S EcC * an bmc (n+m)cC * an bmc (n+m)cck C an bmc (n+m)cck = an bmc (n+m+k+1) (where the first * follows from 4(a), the second from k repetitions of { C cC }) Since (n+m) < (n+m+k+1), L(EcC) L2 L Consider the leftmost productions following S aAE S aAE * aak AE aa k E * aak a n bmc (n+m)c = a(n+k+1)bmc (n+m) (where the first * follows from k repetitions of { A aA }, the second from 4(a)) Since (n+m+k+1) > (n+m), L(aAE) L1 L Consider the leftmost productions following S AU S AU * ak AU a k U * ak a n Ucn a k a n Vcn * ak a n bmVcmc n a k a n bmbBcmc n * ak a n bmbbj Bcmc n ak a n bmbbj c mc n = a(k+n)b(m+j+1)c (m+n) (where the first, second, third and fourth * are k, n, m, and j repetitions of { A aA }, { U aUc }, { V bVc }, and { B bB }, respectively)
2. 2. Since ((k+n)+(m+j+1)) > (m+n), L(AU) L1 L L(G) = L(S) = L(EcC) L(aAE) L(AU) L1 L2 = L Consider w L1, w = an bmc (n+m+k) for some n, m, k, with k > 0 Then w has the leftmost derivation: S EcC * an bmc (n+m)cC * an bmc (n+m)cc(k-1)C an bmc (n+m)cck(k-1) = a n bmc (n+m+k) = w L2 L(G) Consider w L2, w = a(n+j)b(m+k)c (n+m) for some j, k, m, n, with j+k > 0 If k = 0, w has the leftmost derivation: S aAE * aa(j-1)AE aa(j-1)E * aa(j-1)a n Ecn aa(j-1)a n Fcn * aa(j- 1)a n bmFcmc n aa(j-1)a n bmc mc n = a(j+n)b(0+m)c (m+n) = w L1 (k = 0) L(G) If k > 0, then w has the leftmost derivation: S AU * aj AU aj U * aj a n Ucn aj a n Vcn * aj a n bmVcmc n a j a n bmbBcmc n aj a n bmbb(k-1)Bcmc n aj a n bmbb(k-1)c mc n = a(j+n)b(m+k)c (m+n) = w L1 (k > 0) L(G) L1 L(G) L = L1 L2 L(G) L = L(G)